Finding normal force from kinetic and potential energies

In summary: So the correct formula would be V2=sqrt(V1^2 + 2*G*Y). In summary, a 69.2-kg skier with a speed of 8.13 m/s encounters a dip in the snow's surface with a circular cross section of radius 13.2 m and a height difference of 1.63 m. Using the formula for conservation of energy, the skier's final velocity is calculated to be 5.84092 m/s. The normal force exerted by the snow on the skier at point B is determined to be 709.473 N. However, there is a sign error in the calculation for final velocity, as the height at point B should be a
  • #1
lsu777
7
0
A 69.2-kg skier encounters a dip in the snow's surface that has a circular cross section with radius of curvature of r = 13.2 m. If the skier's speed at point A in the figure below is 8.13 m/s, what is the normal force exerted by the snow on the skier at point B?

http://educog.com/res/prenhall/walker/Physics_3E/Chap08/graphics/walk0850.gif"

As shown, h = 1.63 m. Ignore frictional forces.




ok so we are given:
mass=69.2kg
radius=13.2m
Velocity1=8.13 m/s
Change in Y=1.63m
gravity=9.81m/s^2

we know KE1 + PE1=KE2 + PE2

with potential energy2= O


ok so first thing I did was try and find velocity2 with the formula

so 1/2*m*v1^2 + 0= 1/2*m*V2^2 + m*g*Y

so V2=sqrt(V1^2 - 2*G*Y)

I got V2=5.84092


Then I did sum of Forces Y=(normal force - m*g)= mass * Arad


Arad= V2^2/R

so normal force=(mass*(V2^2/R))+m*g

so normal force = 709.473 N



can somebody smarter then me please check this. I think I got it wrong the first time because of math error. ****ing lon-capa sucks and doesn't give you any info. I have one more shot to get it right.
 
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  • #2
I can't see the diagram without a username and password. Is B lower than A, or higher? Is v1 the speed at A, or B? Make sure everything is consistent.
 
  • #3
ideasrule said:
I can't see the diagram without a username and password. Is B lower than A, or higher? Is v1 the speed at A, or B? Make sure everything is consistent.

A is original heigth, B is the bottom of the circle, below A.

V1= Velocity at A


picture was attached.
 

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  • #4
lsu777 said:
so 1/2*m*v1^2 + 0= 1/2*m*V2^2 + m*g*Y

so V2=sqrt(V1^2 - 2*G*Y)

I got V2=5.84092
Is v1 the speed at A, or B? Is Y the y coordinate at A, or B? Your method is correct, but you have a sign issue. (One way to see that you made a mistake is to note that V2 is smaller than the initial speed, which doesn't make sense.)
 
  • #5
ideasrule said:
Is v1 the speed at A, or B? Is Y the y coordinate at A, or B? Your method is correct, but you have a sign issue. (One way to see that you made a mistake is to note that V2 is smaller than the initial speed, which doesn't make sense.)

v1 is speed at A, Y is the height at B to the level of A. can you tell me where I made a mistake on my signs?
 
  • #6
If your height at A is 0, your height at B should be negative, not positive.
 

What is normal force?

Normal force is the force exerted by a surface on an object that is in contact with it. It is always perpendicular to the surface and prevents objects from passing through each other.

How is normal force related to kinetic and potential energies?

Normal force is not directly related to kinetic and potential energies. However, it is an important factor to consider when calculating these energies as it affects the motion and stability of an object.

How do you calculate normal force from kinetic and potential energies?

To calculate normal force from kinetic and potential energies, you first need to determine the total mechanical energy of the system. This includes the object's kinetic energy, potential energy, and any other forms of energy present. Then, using the formula E = 1/2mv^2 + mgh, you can solve for the normal force by rearranging the equation to N = mg + 1/2mv^2.

What are some common examples of normal force in everyday life?

Some common examples of normal force include the force exerted by the ground on our feet when we walk, the force exerted by a table on a book placed on it, and the force exerted by a car's tires on the road as it drives. Normal force is present in any situation where there is contact between two surfaces.

How does normal force affect the motion of an object?

The normal force acts as a reaction force to the weight of the object, and thus, it affects the object's motion by either supporting it or opposing it. If the normal force is greater than the weight, the object will be pushed upwards, and if it is less than the weight, the object will be pushed downwards. In both cases, this affects the object's acceleration and velocity.

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